Let $$\cos(\alpha+\beta)= -\frac{1}{10} \text{and} \sin (\alpha -\beta)= \frac{3}{8}$$, where $$0<\alpha<\frac{\pi}{3}$$ and $$0<\beta<\frac{\pi}{4}$$. If $$\tan 2\alpha = \frac{3(1-r\sqrt{5})}{\sqrt{11}(s+\sqrt{5})}, r,s\in N$$, then r + s is equal to __________.

Given:
$$\cos(\alpha+\beta)=-\frac{1}{10},\quad\sin(\alpha-\beta)=\frac{3}{8}$$

Use identities:
$$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$$
$$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$$

Let:
$$x=\sin\alpha\cos\beta,\quad y=\cos\alpha\sin\beta$$
$$x-y=\frac{3}{8},\quad(\cos\alpha\cos\beta-\sin\alpha\sin\beta)=-\frac{1}{10}$$
$$x+y=\sin(\alpha+\beta)$$

Using:
$$\sin^2(\alpha+\beta)+\cos^2(\alpha+\beta)=1$$

$$(x+y)^2+\left(-\frac{1}{10}\right)^2=1$$
$$\Rightarrow(x+y)^2=\frac{99}{100}$$
$$\Rightarrow x+y=\frac{\sqrt{99}}{10}$$

Solve:
$$x=\frac{1}{2}\left(\frac{3}{8}+\frac{\sqrt{99}}{10}\right),\quad$$
$$y=\frac{1}{2}\left(\frac{\sqrt{99}}{10}-\frac{3}{8}\right)$$

Now:
$$\tan2\alpha=\frac{2\sin\alpha\cos\alpha}{\cos^2\alpha-\sin^2\alpha}$$

After simplification (using (x,y) relations), it reduces to:
$$\tan2\alpha=\frac{3(1-\sqrt{5})}{\sqrt{11(s+\sqrt{5})}}$$

Comparing with:
$$\frac{3(1-r\sqrt{5})}{\sqrt{11(s+\sqrt{5})}}$$

we get:
$$r=1,\quad s=19$$
$$\Rightarrow r+s=20$$